k(k−1)2 is really just the sum of integers from 1 to k-1, and the cosine of some integer multiple π oscillates between -1 and 1 (if the integer is even, it's 1, and if it's odd, it's -1). Here, the sum alternates 2 negatives and 2 positives (starting with k=19)
This means that the first 4 terms of the sum are
−(1+2+3+⋯+18)−(1+2+3+⋯+19)+(1+2+3+⋯+20)+(1+2+3+⋯+21)=20+19+20+21=4(20)
(notice a bunch of cancellation happens between the sums, as the 1+2+3+...18 is repeated throughout all the 4 terms)
So for every group of 4 terms, the sum of the group is the biggest number of the sum of the 3rd term of that group times 4.
Therefore, the sum is:
4(20+24+28+⋯+96)
Use the sum formula to obtain your answer.
We want to compute cosk(k−1)π2 for various intergers k. The first values are as follows:
kcosk(k−1)π2112−13−141516−17−181
The values seem to be periodic with period 4. If we let f(k)=k(k−1)π2, then
f(k+4)−f(k)=(k+4)(k+3)π2−k(k−1)π2=(4k+6)π=(2k+3)⋅2π
Since f(k+4) and f(k) differ by a mutiple of 2π, cosf(k) is indeed period with period of 4. So for a positive integer m,
A4m−1=(4m−1)(4m−2)2cosf(4m−1)=−(4m−1)(4m−2)2,A4m=(4m)(4m−1)2cosf(4m)=(4m)(4m−1)2,A4m+1=(4m+1)(4m)2cosf(4m+1)=(4m+1)(4m)2,A4m+2=(4m+2)(4m+1)2cosf(4m+2)=−(4m+2)(4m+1)2,
Therefore:
A4m−1+A4m+A4m+1+A4m+2=−(4m−1)(4m−2)2+(4m)(4m−1)2+(4m+1)(4m)2−(4m+2)(4m+1)2=−2.
Summing over 5≤m≤24, we find that A19+A20+⋯+A98=20(−2)=−40.